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Math4201Topology I (Lecture 20)

Math4201 Topology I (Lecture 20)

Quotient topology

More propositions

Proposition for quotient maps in restrictions

Let X,YX,Y be topological spaces and p:XYp:X\to Y is surjective and open/closed. Let AXA\subseteq X be saturated by pp, (p1(p(A))=Ap^{-1}(p(A))=A).

Then q:Ap(A)q: A\to p(A) given by the restriction of pp is open/closed surjective map (In particular, it’s a quotient map).

Proof

qq is surjective and continuous. Now assume pp is open and we will show that qq is also open. Any open subspace of AA is given as UAU\cap A where UU is open in XX. By definition, q(UA)=p(UA)=p(U)p(A)q(U\cap A)=p(U\cap A)=p(U)\cap p(A)

To see the second identity:

  1. p(UA)p(U)p(A)p(U\cap A)\subseteq p(U)\cap p(A)

yp(UA)\forall y\in p(U\cap A), y=p(x)y=p(x) with xUAx\in U\cap A, since xAx\in A and xUx\in U, y=p(x)p(U)p(A)y=p(x)\in p(U)\cap p(A)

  1. p(U)p(A)p(UA)p(U)\cap p(A)\subseteq p(U\cap A)

yp(U)p(A)\forall y\in p(U)\cap p(A), y=p(x1)y=p(x_1) with x1Ux_1\in U and y=p(x2)y=p(x_2) with x2Ax_2\in A, since x1Ux_1\in U and x2Ax_2\in A, y=p(x1)=p(x2)p(UA)y=p(x_1)=p(x_2)\in p(U\cap A)

So x1=x2UAx_1=x_2\in U\cap A, y=p(x1)=p(x2)p(U)p(A)y=p(x_1)=p(x_2)\in p(U)\cap p(A), yp(UA)y\in p(U\cap A).

Note that p(U)Xp(U)\subseteq X is open by pp is an open map.

So p(U)p(A)p(U)\cap p(A) is open in p(A)p(A).

q(UA)=p(UA)=p(U)p(A)q(U\cap A)=p(U\cap A)=p(U)\cap p(A) is open.

So qq is open in p(A)p(A).

Simplicial complexes (extra chapter)

Definition for simplicial complexes

Simplicial complexes are topological space with simplices (nn dimensional triangles) as their building blocks.

Definition for n dimensional simplex

Let v0,,vnv_0,\dots,v_n be points in Rm\mathbb{R}^m such that vnv0v_n-v_0, vn1v0v_{n-1}-v_0, \cdots, and v1v0v_1-v_0 are linearly independent in Rm\mathbb{R}^m. (in particular nmn\leq m).

The nn-dimensional simplex determined by {v0,,vn}\{v_0,\dots,v_n\} is given as:

Δn[v0,,vn]={t0v0+t1v1++tnvnti0,i=0nti=1}\Delta^n\coloneqq [v_0,\dots,v_n]=\{t_0v_0+t_1v_1+\cdots+t_nv_n\vert t_i\geq 0, \sum_{i=0}^n t_i=1\}

The coefficients t0,,tnt_0,\dots,t_n are called barycentric coordinates.

Example of simplicial complex

n=0n=0,

Δ0={v0}\Delta^0=\{v_0\}

n=1n=1,

Δ1={t0v0+t1v1t0+t1=1}\Delta^1=\{t_0v_0+t_1v_1\vert t_0+t_1=1\}, this is the line segment between v0v_0 and v1v_1.

n=2n=2,

Δ2={t0v0+t1v1+t2v2t0+t1+t2=1}\Delta^2=\{t_0v_0+t_1v_1+t_2v_2\vert t_0+t_1+t_2=1\}, this is the triangle with vertices v0,v1,v2v_0,v_1,v_2.

Note

Every non-empty subset {vi0,,vik}\{v_{i_0},\dots,v_{i_k}\} of {v0,,vn}\{v_0,\dots,v_n\} determines a kk dimensional simplex [vi0,,vik]Δn=[v0,,vn][v_{i_0},\dots,v_{i_k}]\subseteq \Delta^n=[v_0,\dots,v_n]. Inside the nn dimensional simplex ti0vi0++tinvikΔnt_{i_0}v_{i_0}+\cdots+t_{i_n}v_{i_k}\in \Delta^n. Where the coefficient tjt_j of vj{vi0,,vin}v_j\notin \{v_{i_0},\dots,v_{i_n}\} is 00.

Any such kk dimensional simplex is called a face of the simplex [vi0,,vin][v_{i_0},\dots,v_{i_n}].

Example of faces for simplicial complex

\

For a triangle [v0,v1,v2][v_0,v_1,v_2], the faces are [v0,v1][v_0,v_1], [v0,v2][v_0,v_2], and [v1,v2][v_1,v_2] (the edges of the triangle).

Definition for abstract simplicial complex

Let VV be a finite or countable set, an abstract simplicial complex on VV is a collection of finite non-empty subset of VV, denoted by KK. And the two conditions are satisfied:

  1. If σK\sigma\in K and τσ\tau\subseteq \sigma, then τK\tau\in K.

  2. For any vVv\in V, {v}K\{v\}\in K.

Example of abstract simplicial complex

Let V={a,b,c,d}V=\{a,b,c,d\}.

If we want to include {a,b,c}\{a,b,c\}, then we need to include {a,b}\{a,b\} and {b,c}\{b,c\}, so we have K={{a,b,c},{a,b},{b,c},{a},{b},{c},{d}}K=\{\{a,b,c\},\{a,b\},\{b,c\},\{a\},\{b\},\{c\},\{d\}\} is an abstract simplicial complex.

Topological realization of abstract simplicial complex

Let σKΔσ1\bigsqcup_{\sigma\in K}\Delta^{|\sigma|-1} be the disjoint union of all σ1|\sigma|-1 dimensional simplices in KK.

Xk~=σKΔσ1\tilde{X_k}=\bigsqcup_{\sigma\in K}\Delta^{|\sigma|-1}

We use subspace topology to define a topology on Δn\Delta^n and the union of such topology for each Δσ1\Delta^{|\sigma|-1} defines a topology on Xk~\tilde{X_k}.

We define the equivalence relation xΔσσ1xΔσσ1x\in \Delta_{\sigma}^{|\sigma|-1}\sim x'\in \Delta_{\sigma'}^{|\sigma'|-1} if xΔσσσσ1Δσσ1x\in \Delta_{\sigma'\cap \sigma}^{|\sigma'\cap \sigma|-1}\subseteq \Delta_{\sigma}^{|\sigma|-1}. and xΔσσσσ1Δσσ1x'\in \Delta_{\sigma'\cap \sigma}^{|\sigma'\cap \sigma|-1}\subseteq \Delta_{\sigma'}^{|\sigma'|-1}.

are the sample points of Δσσσσ1\Delta_{\sigma\cap \sigma'}^{|\sigma\cap \sigma'|-1}.

XKX_K is the quotient space of Xk~\tilde{X_k} by the equivalence relation.

Continue next time.

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